nt.number theory - Separable and algebraic closures?

First of all, there is not the algebraic/separable closure. Choices have to be made. However, if an algebraic closure $k^{mathrm{alg}}$ of $k$ is fixed, inside it there is a unique separable closure $k^{mathrm{sep}}$ of $k$, namely the subfield consisting of the separable elements over $k$.

Ignoring the failure of uniqueness, you can regard $k^{mathrm{alg}}$ as the biggest algebraic extension of $k$, whereas $k^{mathrm{sep}}$ is the biggest galois extension of $k$. The latter is because $k^{mathrm{sep}}$ is easily seen to be normal. In particular, you can apply Galois theory and relate the group theory of the absolute Galois group $mathrm{Gal}(k^{mathrm{sep}}/k)$ with the field theory of Galois extensions of $k$. The algebraic closure is too big to make Galois theory work.

Obviously $k$ is perfect if and only if $k^{mathrm{alg}} = k^{mathrm{sep}}$. Finite fields and fields of characteristic $0$ (in particular number fields) are perfect. But what is the difference in the other cases? Let $p = mathrm{char}(k) > 0$. Then $k^{mathrm{alg}} / k^{mathrm{sep}}$ is purely inseparable, i.e. for every $a in k^{mathrm{alg}}$ there is some $n geq 1$ such that $a^{p^n} in k^{mathrm{sep}}$. In other words, this field extension is given by adjoining all $p^n$-th roots. A consequence of this is that the restriction map $mathrm{Aut}_k(overline{k}) to mathrm{Gal}(k^{mathrm{sep}}/k)$ is an isomorphism.

Actually one can show that the canonical map $k^{mathrm{sep}} otimes_k k^{mathrm{perf}} to k^{mathrm{alg}}$ is an isomorphism, where $k^{mathrm{perf}}=cup_{n geq 0} k^{1/p^n}$ is the perfect hull of $k$.